Linear Algebra Examples

$x - 7 y = - 35$ , $- 3 x + 4 y = 5$

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 7 & - 35 \\ - 3 & 4 & 5 \end{array}]$

Row reduce.

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Perform the row operation $R_{2} = 3 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = 3 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - 7 & - 35 \\ 3 \cdot R_{1} + R_{2} & 3 \cdot R_{1} + R_{2} & 3 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = 3 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = 3 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & - 7 & - 35 \\ (3) \cdot (1) - 3 & (3) \cdot (- 7) + 4 & (3) \cdot (- 35) + 5 \end{array}]$

$R_{2} = 3 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - 7 & - 35 \\ 0 & - 17 & - 100 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{17} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - \frac{1}{17} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - 7 & - 35 \\ - \frac{1}{17} R_{2} & - \frac{1}{17} R_{2} & - \frac{1}{17} R_{2} \end{array}]$

$R_{2} = - \frac{1}{17} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{17} R_{2}$ .

$[\begin{array}{c} 1 & - 7 & - 35 \\ (- \frac{1}{17}) \cdot (0) & (- \frac{1}{17}) \cdot (- 17) & (- \frac{1}{17}) \cdot (- 100) \end{array}]$

$R_{2} = - \frac{1}{17} R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - 7 & - 35 \\ 0 & 1 & \frac{100}{17} \end{array}]$

Perform the row operation $R_{1} = 7 \cdot R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = 7 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 7 \cdot R_{2} + R_{1} & 7 \cdot R_{2} + R_{1} & 7 \cdot R_{2} + R_{1} \\ 0 & 1 & \frac{100}{17} \end{array}]$

$R_{1} = 7 \cdot R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = 7 \cdot R_{2} + R_{1}$ .

$[\begin{array}{c} (7) \cdot (0) + 1 & (7) \cdot (1) - 7 & (7) \cdot (\frac{100}{17}) - 35 \\ 0 & 1 & \frac{100}{17} \end{array}]$

$R_{1} = 7 \cdot R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & \frac{105}{17} \\ 0 & 1 & \frac{100}{17} \end{array}]$

Use the result matrix to declare the final solutions to the system of equations.

$x = \frac{105}{17}$

$y = \frac{100}{17}$

The solution is the set of ordered pairs that makes the system true.

$(\frac{105}{17}, \frac{100}{17})$

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